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Let $\rho(t)$ be a function on the set $\mathbb{R}^+$ of nonnegative real numbers such that:

  • $\rho$ is nondecreasing (and continuous - thanks for the correction)
  • $\rho(t) = 0$ if and only if $t = 0$

Let $X$ be a metric space and let $f$ be a real valued function on $X$. Say that $f$ has modulus of continuity $\rho$ if $|f(x) - f(y)| \leq \rho(d(x,y))$ for every $x$ and $y$ in $X$. For example, a function is Lipschitz if and only if it has modulus of continuity $Ct$ for some positive real number $C$. Observe that a function with modulus of continuity $\rho$ is necessarily continuous.

Question: If $X$ is a compact metric space without isolated points, is it true that the set of all functions with modulus of continuity $\rho$ is nowhere dense (meaning its closure contains no open set) in $C(X)$ equipped with the supremum norm?

I am a TA in a class in which it was claimed that the answer is yes, but I don't completely believe the proof given and I can't seem to find a correct argument except in special cases. For example, one can show that the set of all Lipschitz functions on $[0,1]$ with Lipschitz constant $C$ is nowhere dense in $C[0,1]$ using the existence of piecewise linear functions of arbitrarily small norm whose linear pieces all have slope larger than $C$ (or smaller than -C). So the idea for general $X$ should be to construct continuous functions of arbitrarily small norm with arbitrarily rapid oscillation, but I don't see how to do this.

Thanks!

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$f$ is necessarily continuous only if $\rho$ is continuous, no? –  Thomas Andrews Nov 16 '11 at 21:29
    
Or rather, you need continuity at 0, or, since you defined $\rho$ only on $\mathbb R^+$, you need $\rho(x)\to 0$ as $x\to 0$ –  Thomas Andrews Nov 16 '11 at 21:41
    
Quite right, I'll edit. –  Paul Siegel Nov 16 '11 at 22:12
    
The set of functions that have modulus of continuity $\rho$ at $x_0$ is not a set of continuous functions - the condition does not say anything about continuity at points $x\neq x_0$ –  Thomas Andrews Nov 17 '11 at 16:15
    
Good point, I edited again (sorry it's taking me so long to get it right). I want to fix a modulus $\rho$ and look at the set of all continuous functions which have modulus of continuity $\rho$ at some point of $X$. Basically the point is to generalize the crazy piecewise linear example given above to an arbitrary metric space. –  Paul Siegel Nov 17 '11 at 18:32

2 Answers 2

up vote 1 down vote accepted

Let $C^\rho(X,x_0)$ be the set of continuous functions $X\to \mathbb R$ with modulus of continuity $\rho$ at $x_0$.

Observation (1): $C^\rho(X,x_0)$ is closed in $C(X)$. If $f$ is in the closure, then: $$|f(x_0)-f(x)|\leq|f(x_0)-g(x_0)| + |g(x_0)-g(x)| + |g(x)-f(x)|\leq \rho(d(x_0,x)) + 2\sup_{z\in X} |f(z)-g(z)|$$

Where $g\in C^\rho(X,x_0)$ But we can make $\sup_{z\in X} |f(z)-g(z)|$ be arbitrarily small since $f$ is in the closure, so $|f(x_0)-f(x)|\leq \rho(d(x_0,x))$.

Observation (2): If $f,g\in C^\rho(X,x_0)$, then $f-g\in C^{2\rho}(X,x_0)$. This is easy to see.

So, if, for every $\epsilon>0$, we can find an $h\in C(X)$ such that $|h(x)|<\epsilon$ for all $x\in X$ and $h \notin C^{2\rho}(X,x_0)$, then you are done, because for any $f\in C^\rho(X,x_0)$, $f+h\notin C^\rho(X,x_0)$, and therefore $C^\rho(X,x_0)$ is nowhere dense in $C(X)$.

Given $\epsilon>0$, we pick an $x_1\neq x_0$ so that $4\rho(d(x_0,x_1))<\epsilon$. You can find such $x_1$ since $x_0$ is not an isolated point and $\rho(t)\to 0$ as $t\to 0$. Define $\delta=d(x_0,x_1)>0$.

Define $\phi(t)=\frac{\epsilon}{2}(1-\frac{t}{\delta})$ if $t\leq \delta$ and $\phi(t)=0$ if $t>\delta$. Then $h(x)=\phi(d(x_0,x))$ has the property that $|h(x)|<\epsilon$, $h(x_0)=\frac{\epsilon}2$, and $h(x_1)=0$. So $|h(x_0)-h(x_1)|=\frac{\epsilon}2>2\rho(d(x_0,x_1))$. So $h(x)\notin C^{2\rho}(X,x_0)$

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  • How to construct greatly oscillating functions with small uniform norms :

well in this case you only need a quick oscillation "at one point", together with arbitrary small norm.

Choose 2 points $a$ and $b$ in $X$ with $d(a,b)< \epsilon$. It is possible since $X$ has no isolated points. By Urysohn's lemma there is a continuous function $h : X \rightarrow [0,1]$ such that $h(a)=1$ and $h(b)=0$. Now let $g$ be the continuous function on $X$ defined by $g(x)=\alpha h(x)$, and $f$ such that $f$ admits $\rho$ as a modulus. For $\alpha$ arbitrarily small, $f+g$ belongs to $B(f,\alpha)$ but for $\epsilon$ small enough it will not admit $\rho$ as a continuity modulus.

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Can you make your construction work at every point? (See my edits since your answer was posted). –  Paul Siegel Nov 17 '11 at 18:34
    
well unless I'm mistaken I think this answers your question (I don't see why you'd need a function oscillating quickly everywhere : the set of continuous functions admitting $\rho$ as a modulus of continuity is closed, and my construction shows that it cannot contain any ball, hence it cannot contain any open set). Now if you really want something oscillating violently everywhere, I guess something along the lines of $f(x)=\alpha \sin(d(x,x_0)/\epsilon)$ would do the trick, where $x_0$ is arbitrary (choosing $\epsilon$ small enough and using the fact that $X$ has no isolated points). –  Glougloubarbaki Nov 21 '11 at 19:49

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